Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $q \neq 0$. $p = \dfrac{-3q + 9}{q - 2} \div \dfrac{-3q^2 + 18q + 81}{q^2 - 11q + 18} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $p = \dfrac{-3q + 9}{q - 2} \times \dfrac{q^2 - 11q + 18}{-3q^2 + 18q + 81} $ First factor out any common factors. $p = \dfrac{-3(q - 3)}{q - 2} \times \dfrac{q^2 - 11q + 18}{-3(q^2 - 6q - 27)} $ Then factor the quadratic expressions. $p = \dfrac {-3(q - 3)} {q - 2} \times \dfrac {(q - 9)(q - 2)} {-3(q - 9)(q + 3)} $ Then multiply the two numerators and multiply the two denominators. $p = \dfrac {-3(q - 3) \times (q - 9)(q - 2) } {(q - 2) \times -3(q - 9)(q + 3) } $ $p = \dfrac {-3(q - 9)(q - 2)(q - 3)} {-3(q - 9)(q + 3)(q - 2)} $ Notice that $(q - 9)$ and $(q - 2)$ appear in both the numerator and denominator so we can cancel them. $p = \dfrac {-3\cancel{(q - 9)}(q - 2)(q - 3)} {-3\cancel{(q - 9)}(q + 3)(q - 2)} $ We are dividing by $q - 9$ , so $q - 9 \neq 0$ Therefore, $q \neq 9$ $p = \dfrac {-3\cancel{(q - 9)}\cancel{(q - 2)}(q - 3)} {-3\cancel{(q - 9)}(q + 3)\cancel{(q - 2)}} $ We are dividing by $q - 2$ , so $q - 2 \neq 0$ Therefore, $q \neq 2$ $p = \dfrac {-3(q - 3)} {-3(q + 3)} $ $ p = \dfrac{q - 3}{q + 3}; q \neq 9; q \neq 2 $